2.1.6 · D3 · HinglishAnalytical Mechanics
Worked examples — Applying E-L equations to various systems
2.1.6 · D3· Physics › Analytical Mechanics › Applying E-L equations to various systems
Prerequisites jo tumhare paas already hone chahiye: 2.1.05-Deriving-the-Euler-Lagrange-equation (jahan se recipe aayi hai), Constraints-and-degrees-of-freedom (coordinates kitne chahiye yeh kaise count karein), aur — "conserved quantity" wale cells ke liye — 2.1.07-Cyclic-coordinates-and-conservation-laws.
Ek rule jo hum har example mein use karte hain
Kisi bhi system ko touch karne se pehle, tool ko wall pe pin kar lete hain taaki koi bhi example yeh assume na kare ki tumhe yeh yaad hai.
Neeche har example sirf yahi rule hai, apply kiya gaya. "Forecast:" line padho aur scroll karne se pehle guess karo.
Scenario matrix
Har E–L problem in case classes mein se kisi ek mein fit hota hai. Columns hain woh tarah ki mushkilein jo ek problem chhupa sakta hai; neeche ke examples tagged hain us cell ke saath jo wo cover karte hain.
| Cell | Case class | Kya tricky banaata hai | Kis Example mein covered |
|---|---|---|---|
| A | 1 DOF, clean | bas recipe follow karo | Ex 1 |
| B | Sign / direction bookkeeping | "positive" kaunsi direction hai? PE ke signs | Ex 2 |
| C | Degenerate input (mass, length, ya angle ) | formula limit mein survive kare | Ex 3 |
| D | Limiting / extreme geometry (angle , ) | known physics se check karo | Ex 4 |
| E | Non-conservative / driven (external applied force) | plain kaafi nahi | Ex 5 |
| F | Coupled 2 DOF (do equations, cross-terms) | aur entangle ho jaate hain | Ex 6 |
| G | Cyclic coordinate → conserved momentum | woh coordinate dhundho jo mein absent hai | Ex 7 |
| H | Real-world word problem | words ko mein translate karo | Ex 8 |
| I | Exam-style twist (moving support / time in ) | explicitly appear karta hai | Ex 9 |
Hum nau cells se nau examples ke saath guzarenge.
Example 1 — clean 1-DOF baseline · Cell A
Step 1 — DOF count karo. Yeh step kyun? Sab kuch is sawaal se start hota hai: "system ko locate karne ke liye mujhe kitne independent numbers chahiye?" Block ek seedhi rail pe move karta hai ⇒ 1 DOF, coordinate .
Step 2 — likho. Kyun? Kinetic energy hai aur yahan :
Step 3 — likho. Kyun? Ek stretched/compressed spring store karta hai — standard elastic potential:
Step 4 — banao aur uske do partial derivatives. Kyun do? E–L rule ko (the "momentum slot") aur (the "force slot") dono chahiye.
\frac{\partial L}{\partial\dot x}=m\dot x,\qquad \frac{\partial L}{\partial x}=-kx.$$ **Step 5 — E–L mein plug karo.** *Sirf pehle term pe $d/dt$ kyun?* Kyunki rule hai $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot x}-\dfrac{\partial L}{\partial x}=0$: $$m\ddot x-(-kx)=0\;\Rightarrow\;\boxed{m\ddot x=-kx}.$$ > [!formula] Result > $$\ddot x=-\frac{k}{m}\,x,\qquad \omega=\sqrt{\frac{k}{m}}.$$ **Verify:** $m=2\ \text{kg}$, $k=8\ \text{N/m}$ ke saath: $\omega=\sqrt{8/2}=2\ \text{rad/s}$. Units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{\text{s}^{-2}}=\text{s}^{-1}$ ✔. Yeh exactly Newton's $F=-kx$ hai — baseline kaam kar raha hai. --- ## Example 2 — sign bookkeeping galat phir sahi · Cell B > [!example] Do beads, ek pulley, "dangerous" tarike se choose kiya > Atwood machine ($m_1,m_2$, frictionless pulley, inextensible string) dobara karo lekin is baar $y$ measure karo ki $m_2$ kitna **upar** utha. $m_2$ ka acceleration nikalo. > > **Forecast:** kya answer ka sign parent note ke version se flip hoga (jo "$m_1$ descends" use karta tha)? Guess karo yes/no. **Step 1 — DOF & coordinate.** *Ek coordinate kyun?* String do masses ko couple karti hai: ek fix karo toh doosra determined hai ⇒ **1 DOF**. Humara chosen $y$ = $m_2$ ki rise. Tab $m_1$ usi $y$ se **neeche** girta hai (string length fixed). **Step 2 — $T$.** *Dono $\dot y$ se kyun move karte hain?* Inextensibility equal speeds force karti hai: $$T=\tfrac12 m_1\dot y^2+\tfrac12 m_2\dot y^2=\tfrac12(m_1+m_2)\dot y^2.$$ **Step 3 — $V$ (sign minefield).** *Yahan sign care kyun matter karti hai:* PE **badhti** hai jab mass **upar** jaata hai. $m_2$ $y$ se upar jaata hai ⇒ uski PE $+m_2 g y$ hai. $m_1$ $y$ se neeche girta hai ⇒ uski PE $-m_1 g y$ hai: $$V=m_2 g y-m_1 g y=(m_2-m_1)g\,y.$$ **Step 4 — Partial derivatives.** *Yeh do kyun:* momentum slot $\partial L/\partial\dot y$ bahar wale $d/dt$ ko feed karta hai, aur force slot $\partial L/\partial y$ usse subtract hota hai, rule ke according. $$\frac{\partial L}{\partial\dot y}=(m_1+m_2)\dot y,\qquad \frac{\partial L}{\partial y}=-(m_2-m_1)g=(m_1-m_2)g.$$ **Step 5 — E–L mein assemble karo.** *Pehle slot ko time mein differentiate kyun karte hain:* rule kehta hai rate-of-change of momentum minus force $=0$; yahan $\frac{d}{dt}(m_1+m_2)\dot y=(m_1+m_2)\ddot y$: $$(m_1+m_2)\ddot y-(m_1-m_2)g=0\;\Rightarrow\;\boxed{\ddot y=\frac{(m_1-m_2)g}{m_1+m_2}}.$$ **Verify:** Agar $m_1>m_2$ (bhaari mass neeche girta hai) toh answer **positive** hai ⇒ $m_2$ *upar* ki taraf accelerate karta hai — exactly wahi jo physical sense demand karta hai. $m_1=3,\,m_2=1,\,g=9.8$ ke saath: $\ddot y=\frac{2}{4}\cdot 9.8=4.9\ \text{m/s}^2$. Parent ke result ke barabar magnitude — physics coordinate-independent hai, sirf *sign ki interpretation* badli. ✔ --- ## Example 3 — degenerate input: ek mass zero ho jaata hai · Cell C > [!example] Shrinking bob wala pendulum > Simple pendulum EOM $\ddot\theta=-\dfrac{g}{\ell}\sin\theta$ lo. Kya hoga jab $m\to 0$? Jab $\ell\to 0$? Kya machine break ho jaati hai? > > **Forecast:** guess karo ki kya $m\to0$ motion ko change karta hai. **Step 1 — Dekho $m$ kahan baitha hai.** *Wapas kyun dekhein?* Limit lene se pehle dekho ki woh quantity *final law mein appear bhi karti hai ya nahi*: $$L=\tfrac12 m\ell^2\dot\theta^2-mg\ell(1-\cos\theta).$$ Har term mein $m$ ka ek factor hai. **Step 2 — E–L equation se factor out karo.** *Yeh key move kyun hai:* E–L equation ek sum of terms ko zero ke barabar set karti hai, aur **har** term mein exactly ek factor of $m$ hota hai. $m\cdot(\text{stuff})=0$ ki form ki equation, jab $m\neq0$ ho, $(\text{stuff})=0$ ke *equivalent* hai — dono sides ko ek hi nonzero number se divide karna kabhi bhi solutions nahi badalta. Isi liye hum limits ke saath kuch karne se pehle $m$ cancel karne ki *permission* rakhte hain: $$\frac{d}{dt}(m\ell^2\dot\theta)+mg\ell\sin\theta=0 \;\xrightarrow{\ \div\, m\ (m\neq0)\ }\; \ell^2\ddot\theta+g\ell\sin\theta=0.$$ Toh **$m$ EOM se gayab ho gaya** har $m>0$ ke liye. **Step 3 — $m\to 0$ limit.** *Kyun subtle hai:* $m=0$ exactly matlab hai $L\equiv0$ — koi dynamics defined nahi. Lekin *limit* $m\to0^+$ $\ddot\theta=-\frac{g}{\ell}\sin\theta$ ko unchanged rehne deta hai. Physics: ek halka bob *same rate* se swing karta hai (Galileo ki insight). **Step 4 — $\ell\to 0$ limit.** *Kyun dangerous hai:* $\ell$ poori tarah cancel **nahi** hota — yeh do terms mein *alag powers* pe appear karta hai ($\ell^2$ aur $\ell^1$). Step 2 ki equation se start karo, $\ell^2\ddot\theta+g\ell\sin\theta=0$. Dono terms mein **ek** common factor $\ell$ hai, aur $\ell\neq0$, toh usi "nonzero number se divide karo" logic se hum dono sides ko $\ell^2$ se divide karte hain (equivalently: $\ell$ nikalo, cancel karo, phir $\ddot\theta$ isolate karo): $$\ell^2\ddot\theta+g\ell\sin\theta=0 \;\xrightarrow{\ \div\,\ell^2\ (\ell\neq0)\ }\; \ddot\theta+\frac{g}{\ell}\sin\theta=0 \;\Rightarrow\; \boxed{\ddot\theta=-\frac{g}{\ell}\sin\theta}.$$ Ab limit lo: $\ \ell\to0^+\ \Rightarrow\ \ddot\theta\to-\infty$. Bahut chhota pendulum infinitely fast oscillate karta hai ($\omega=\sqrt{g/\ell}\to\infty$). Formula survive karta hai — bas ek extreme predict karta hai. **Verify:** $\omega=\sqrt{g/\ell}$. $\ell=1\ \text{m},\,g=9.8$ ke liye: $\omega=\sqrt{9.8}\approx3.130\ \text{rad/s}$. $\ell=0.01\ \text{m}$ ke liye: $\omega=\sqrt{980}\approx31.30\ \text{rad/s}$ — das guna faster, $\omega\propto\ell^{-1/2}$ ke saath match karta hai. ✔ --- ## Example 4 — limiting geometry: incline angle apne extremes pe · Cell D > [!example] Incline, bilkul flat se vertical tak > Ek block ek frictionless incline of angle $\alpha$ pe slide karta hai. **Coordinate convention:** $s$ ko woh distance maano jo **slope ke neeche** travel ki gayi, positive down-slope direction mein, $s=0$ se start karke. Is choice ke saath along-slope EOM hai $\ddot s=g\sin\alpha$ (parent Ex 3, same down-slope-positive convention). Do **limits** check karo $\alpha\to0°$ aur $\alpha\to90°$, aur beech mein $\ddot s$ ka sign. > > **Forecast:** $\alpha=90°$ pe $\ddot s$ kya hona chahiye? Padhne se pehle bolo. Neeche ki figure setup dikhati hai: ek right triangle jiska hypotenuse slope hai. **Burnt-orange arrow** $s$ labeled *down the slope* point karta hai (hamari positive direction); **teal arrow** resulting vertical drop $s\sin\alpha$ dikhata hai; **plum arc** base pe incline angle $\alpha$ mark karta hai. ![[deepdives/dd-physics-2.1.06-d3-s01.png]] **Step 1 — Figure se geometry padho.** *Yahan figure kyun?* Poora result ek right triangle pe tikaa hai: $s$ ki down-slope slide ke liye (orange arrow), vertical drop angle $\alpha$ ki **opposite** side hai, yaani $s\sin\alpha$ (teal arrow). Kyunki $s$ positive *down-slope* hai, positive $\ddot s$ matlab "downhill accelerate ho raha hai." **Step 2 — Flat limit $\alpha\to 0°$.** *Check kyun karein:* zero slope zero acceleration deni chahiye. $\sin 0°=0\Rightarrow\ddot s=0$ — block bas baitha rehta hai. ✔ **Step 3 — Vertical limit $\alpha\to 90°$.** *Check kyun karein:* vertical "incline" free fall hai. $\sin90°=1\Rightarrow\ddot s=g$ — exactly free fall, aur kyunki $s$ down-slope point karta hai (ab seedha neeche), positive matlab girna. ✔ **Step 4 — Beech mein, sign.** *Sign kyun matter karta hai:* $\sin\alpha>0$ har $0°<\alpha<90°$ ke liye, toh $\ddot s>0$ throughout — hamare down-slope-positive convention ke saath block hamesha *down-slope accelerate* karta hai, kabhi upar nahi. Koi aise quadrant nahi hain jahan formula sign flip kare kyunki ek real incline sirf $(0°,90°)$ mein rehta hai. **Verify:** $\alpha=30°$: $\ddot s=g\cdot0.5=4.9\ \text{m/s}^2$. $\alpha=90°$: $\ddot s=9.8\ \text{m/s}^2$. Units: $g\sin\alpha=\text{m/s}^2\times(\text{dimensionless})=\text{m/s}^2$ ✔. --- ## Example 5 — driven / non-conservative system · Cell E > [!example] Spring pe push kiya gaya cart > Example-1 block ($m$, spring $k$) ko ab ek external, time-dependent horizontal force $F(t)$ push kar rahi hai (maano tumhare haath se). Yeh force **kisi potential se derive nahi hoti**. EOM nikalo. > > **Forecast:** $F(t)$ kahan enter karta hai — $T$ mein, $V$ mein, ya kisi naye jagah? **Step 1 — Plain $L=T-V$ kyun kaafi nahi.** *Kyun:* $L=T-V$ sirf woh forces capture karta hai jo kisi stored potential se aati hain. Haath ka push koi potential nahi rakhta; humein E–L ka **generalized-force** form chahiye: $$\frac{d}{dt}\frac{\partial L}{\partial\dot q}-\frac{\partial L}{\partial q}=Q_q,$$ jahan $Q_q$ non-conservative generalized force hai $q$ slot mein. *Yeh standard extension hai* — left side still $L=T-V$ conservative parts ka hai. **Step 2 — $L$ sirf conservative parts se banao.** $T=\tfrac12 m\dot x^2$, $V=\tfrac12 k x^2$: $$L=\tfrac12 m\dot x^2-\tfrac12 k x^2.$$ **Step 3 — $Q_x$ nikalo.** *Yeh $F$ ke barabar kyun hai:* $Q_x$ woh work hai jo $x$ ke unit displacement pe hoti hai. Kyunki $F$ *along* $x$ act karta hai, $Q_x=F(t)$ directly (koi projection needed nahi). **Step 4 — Assemble karo.** $\dfrac{\partial L}{\partial\dot x}=m\dot x$, $\dfrac{\partial L}{\partial x}=-kx$: $$m\ddot x-(-kx)=F(t)\;\Rightarrow\;\boxed{m\ddot x+kx=F(t)}.$$ **Verify:** Yeh ordinary Newtonian mechanics ka driven-oscillator equation hai ($ma = -kx + F$) ✔. Drive off kar do, $F=0$, aur yeh Example 1 pe collapse ho jaata hai. $m=2,\,k=8,\,F=4\ \text{N}$ constant ke saath, equilibrium shift $x_{\text{eq}}=F/k=4/8=0.5\ \text{m}$ hai ✔. --- ## Example 6 — coupled two-DOF system · Cell F > [!example] Double-spring chain (do masses, do springs) > Masses $m$ aur $m$ frictionless rail pe. Wall–spring($k$)–mass$_1$–spring($k$)–mass$_2$ (free end). $x_1,x_2$ equilibrium se displacements hain. Dono EOMs nikalo. > > **Forecast:** kya har mass ki equation sirf *apne coordinate* mein hogi, ya doosre ki bhi? Neeche ki figure chain dikhati hai: left pe wall, phir ek spring (teal) pehle mass tak (orange dot, displacement $x_1$), phir ek middle spring (plum) doosre mass tak (orange dot, displacement $x_2$). Key visual yeh hai ki **middle spring $x_2-x_1$ difference se stretch hota hai**, kisi ek displacement se nahi. ![[deepdives/dd-physics-2.1.06-d3-s02.png]] **Step 1 — DOF.** *Kyun do:* do masses slide karne ke liye free hain ⇒ **2 DOF**, coordinates $x_1,x_2$ (figure mein do orange dots). **Step 2 — $T$.** *Yeh step kyun?* Humein chosen coordinates mein kinetic energy chahiye; do masses independently move karte hain toh unki kinetic energies simply add ho jaati hain: $$T=\tfrac12 m\dot x_1^2+\tfrac12 m\dot x_2^2.$$ **Step 3 — $V$ (jahan coupling chhupa hai).** *$(x_2-x_1)$ kyun:* middle spring **dono displacements ke difference** se stretch hota hai, kisi ek se nahi (figure mein plum spring dekho): $$V=\tfrac12 k x_1^2+\tfrac12 k (x_2-x_1)^2.$$ **Step 4 — Partial derivatives, ek per coordinate.** *Ek E–L equation har coordinate ke liye kyun:* rule **alag alag** har degree of freedom ke liye fire karta hai, toh $x_1$ aur $x_2$ dono ke liye momentum slot aur force slot chahiye. $$\frac{\partial L}{\partial\dot x_1}=m\dot x_1,\quad \frac{\partial L}{\partial x_1}=-kx_1+k(x_2-x_1),$$ $$\frac{\partial L}{\partial\dot x_2}=m\dot x_2,\quad \frac{\partial L}{\partial x_2}=-k(x_2-x_1).$$ **Step 5 — Do E–L equations.** *Har momentum slot ko time mein differentiate kyun karein:* $\frac{d}{dt}(m\dot x_i)=m\ddot x_i$, phir har force slot subtract karo: $$\boxed{m\ddot x_1=-2k x_1+k x_2},\qquad \boxed{m\ddot x_2=k x_1-k x_2}.$$ *"Coupled" kyun:* $x_2$ mass-1 ki equation mein hai aur vice versa — woh independently move nahi kar sakte, exactly wahi jo shared middle spring karna chahiye. **Verify:** Ek coupled chain ki ek genuine motion **normal mode** hoti hai — ek pattern $x_1=A_1\cos\omega t,\ x_2=A_2\cos\omega t$ jo single frequency pe oscillate karta hai. $\ddot x_i=-\omega^2 x_i$ substitute karne se do boxed equations $-m\omega^2 A_1=-2kA_1+kA_2$ aur $-m\omega^2 A_2=kA_1-kA_2$ mein convert ho jaati hain. Is $2\times2$ system ka determinant zero set karne pe $m^2\omega^4-3km\omega^2+k^2=0$ milta hai, yaani $\omega^2=\frac{k}{2m}(3\pm\sqrt5)$ — do real positive frequencies, jaisi ek stable coupled system mein honi chahiye. Symmetry check: labels swap karna *symmetry nahi* hai kyunki sirf mass 1 wall ko touch karta hai, aur indeed do equations asymmetric hain ✔. --- ## Example 7 — cyclic coordinate → conserved momentum · Cell G > [!example] Free particle horizontal plane mein slide karta hua (koi potential nahi) > Ek puck freely horizontal plane mein slide karta hai, koi forces nahi, polar coordinates $(r,\theta)$ mein describe kiya. Conserved quantity **puri motion solve kiye bina** nikalo. > > **Forecast:** derive karne se pehle conserved quantity ka naam lo. **Step 1 — Polars mein free particle ka $L$.** *Polars kyun:* yeh hidden symmetry expose karte hain. Parent se, $T=\tfrac12 m(\dot r^2+r^2\dot\theta^2)$ aur $V=0$: $$L=\tfrac12 m(\dot r^2+r^2\dot\theta^2).$$ **Step 2 — Missing coordinate pakdo.** *Yeh poori trick kyun hai:* $\theta$ **$L$ mein appear nahi karta** (sirf $\dot\theta$ karta hai). Ek coordinate jo $L$ mein absent ho usse **cyclic** kehte hain; dekho [[2.1.07-Cyclic-coordinates-and-conservation-laws]]. **Step 3 — $\theta$ ke liye E–L.** *Conservation kyun deta hai:* force slot $\partial L/\partial\theta=0$ hai, toh rule $\frac{d}{dt}(\partial L/\partial\dot\theta)=0$ tak collapse ho jaata hai — momentum slot time mein constant rehta hai: $$\frac{\partial L}{\partial\dot\theta}=mr^2\dot\theta\;\Rightarrow\;\boxed{p_\theta=mr^2\dot\theta=\text{const}}.$$ Yeh hai **angular momentum** — symmetry ke through free se conserved, [[Noethers-theorem]] se juda (har continuous symmetry ⇒ ek conservation law). **Verify (numeric, straight-line free motion).** Ek free puck constant speed $v$ pe straight line mein move karta hai. Maano origin se uski perpendicular distance (impact parameter) $b$ hai. Polars mein angular momentum per unit mass $r^2\dot\theta$ hai, aur geometry kehti hai $r^2\dot\theta = r\cdot(r\dot\theta) = r\cdot v_\perp$, jahan $v_\perp$ tangential speed hai. Closest approach pe $r=b$ aur poori velocity tangential hai, toh $v_\perp=v$ deta hai $r^2\dot\theta=bv$. Door $r=2b$ pe, tangential component $v_\perp = v\cdot\frac{b}{2b}=\frac{v}{2}$ ho jaata hai, toh $r^2\dot\theta=(2b)\cdot\frac{v}{2}=bv$ — **wohi value** $bv$. $b=1,\,v=1$ lo: dono $1$ dete hain. Constant confirm hua ✔. --- ## Example 8 — real-world word problem · Cell H > [!example] Playground pe bachche ki slide > Ek bachcha (mass $m=20\ \text{kg}$) ek straight, frictionless slide ke top pe rest se start karta hai jiska length $L=3\ \text{m}$ aur inclination $\alpha=40°$ hai. E–L use karke, bottom pe uski speed kitni hogi? $g=9.8\ \text{m/s}^2$ lo. > > **Forecast:** compute karne se pehle speed estimate karo (m/s mein). **Step 1 — Words ko coordinate mein translate karo.** *Kyun:* "straight slide" ⇒ 1 DOF, $s$ = slope ke **neeche** travel ki distance (same down-slope-positive convention as Example 4), $s=0$ se start. **Step 2 — $T,V$.** *Kyun:* chosen coordinate mein do energies banao. $T=\tfrac12 m\dot s^2$; height drop $=s\sin\alpha$ toh $V=-mg s\sin\alpha$. **Step 3 — EOM (Cell D result reuse karo).** *Reuse kyun:* yeh same incline geometry hai, toh E–L machine same law deta hai: $$\ddot s=g\sin\alpha=9.8\sin40°.$$ $\sin40°\approx0.6428$, toh $\ddot s\approx6.299\ \text{m/s}^2$ (constant). **Step 4 — Exit speed nikalo.** *Ab kinematics kyun:* rest se distance $L$ pe constant acceleration: $v^2=2\,\ddot s\,L$: $$v=\sqrt{2(6.299)(3)}=\sqrt{37.79}\approx6.147\ \text{m/s}.$$ **Verify (energy cross-check):** Energy kehti hai $\tfrac12 m v^2=mg L\sin\alpha\Rightarrow v=\sqrt{2gL\sin\alpha}=\sqrt{2(9.8)(3)(0.6428)}=\sqrt{37.80}\approx6.148\ \text{m/s}$ ✔. Notice karo $m$ cancel ho gaya — ek bhaari bachcha *same speed* pe pahunchta hai. Units: $\sqrt{\text{m/s}^2\cdot\text{m}}=\text{m/s}$ ✔. --- ## Example 9 — exam twist: time explicitly appear karta hai (moving support) · Cell I > [!example] Pendulum jiska pivot constant speed pe sideways drag kiya jaata hai > Ek simple pendulum (length $\ell$, bob $m$) ka pivot horizontally $X(t)=ut$ (constant speed $u$) ke roop mein move karne ke liye force kiya jaata hai. Coordinate: swing angle $\theta$. EOM nikalo. > > **Forecast:** kya ek *constant-velocity* moving pivot pendulum ki equation badalta hai? Guess karo. **Step 1 — Bob ki position.** *Pehle Cartesian mein kyun:* moving pivot ke saath velocity mein do contributions hoti hain jinhe hum carefully add karein. Pivot ke saath $(ut,0)$ pe: $$x=ut+\ell\sin\theta,\qquad y=-\ell\cos\theta.$$ **Step 2 — Velocities.** *Dono terms differentiate kyun karein:* pivot drift $u$ swing mein add hota hai: $$\dot x=u+\ell\dot\theta\cos\theta,\qquad \dot y=\ell\dot\theta\sin\theta.$$ **Step 3 — $T$.** Square karo aur add karo. *Cross term kyun appear hota hai:* $\dot x^2$ expand karne se $u$ aur $\ell\dot\theta\cos\theta$ mix ho jaate hain: $$T=\tfrac12 m\big(u^2+2u\ell\dot\theta\cos\theta+\ell^2\dot\theta^2\big).$$ **Step 4 — $V$ aur $L$.** $V=mg y=-mg\ell\cos\theta$: $$L=\tfrac12 m u^2+m u\ell\dot\theta\cos\theta+\tfrac12 m\ell^2\dot\theta^2+mg\ell\cos\theta.$$ **Step 5 — $\theta$ ke liye E–L.** *Cross term ke saath careful kyun rahein:* $$\frac{\partial L}{\partial\dot\theta}=mu\ell\cos\theta+m\ell^2\dot\theta,\qquad \frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot\theta}\right)=-mu\ell\dot\theta\sin\theta+m\ell^2\ddot\theta.$$ $$\frac{\partial L}{\partial\theta}=-mu\ell\dot\theta\sin\theta-mg\ell\sin\theta.$$ Subtract karo: do $-mu\ell\dot\theta\sin\theta$ terms **cancel** ho jaate hain: $$m\ell^2\ddot\theta+mg\ell\sin\theta=0\;\Rightarrow\;\boxed{\ddot\theta=-\frac{g}{\ell}\sin\theta}.$$ **Verify:** Stationary pendulum se *bilkul same*! *Kyun aisa hona chahiye:* **constant velocity** pe move karta pivot sirf ek inertial frame hai (Galilean relativity) — koi naya force nahi. Agar $X(t)=\tfrac12 a t^2$ hota (accelerating pivot), toh cross term ek extra $-\frac{a}{\ell}\cos\theta$ chhod jaata aur equation badal jaati. Toh yeh twist sikhata hai: sirf support ka *acceleration* matter karta hai. ✔ --- ## Active recall > [!recall]- Kaun sa cell kaun sa hai? (answers cover karo) > - Mass ya length $\to 0$ aur formula still holds — kaun sa cell? ::: Cell C (degenerate input) > - Ek hand-push jiska koi potential nahi — kaun sa cell, kaun sa extra symbol? ::: Cell E; generalized force $Q_q$ > - $\theta$ $L$ mein missing hone se kya milta hai? ::: ek conserved momentum $p_\theta$ (Cell G) > - Do coordinates ek doosre ki equations mein appear karein — iska naam kya hai? ::: coupled system (Cell F) > - Constant-velocity moving pivot pendulum EOM badalta hai? ::: Nahi — inertial frame; sirf support ka acceleration matter karta hai (Cell I) > [!mnemonic] Nau cells > **A**–clean, **B**–signs, **C**–zero, **D**–extremes, **E**–external force, **F**–coupled, **G**–cyclic, **H**–word, **I**–time. *"A Bad Cat Dashed, Every Fluffy Gato Hid Inside."*